D＝√（4P）
/
（πτ）
＝√
/
（4×800）
（3.14×19.2）
1271-1272
d
h
r
b
d
r
＝2.4674Dd2
c
M
3D shape
＝πth（D－t）
＝πth（d＋t）
b
πh
V＝ 6 （3a 2＋3b 2＋h 2）
A,a＝Surface area of
each end
P
D
d
When curve has
circumference that is an arc:
πℓ
V＝ 12（2D 2＋d 2）
When curve has
circumference that is a
parabola
V＝0.209ℓ（2D2Dd＋1/4d2）
ℓ
d
4
V＝ 3 πr3＝4.1888r3
π
＝ 6 d3＝0.5236d3
■Physical properties of metal materials
Density Young´s modulus E Coefficient of thermal expansion
［g/cm3］ ［kgf/mm2］
［M10-6 /°C］
Soft steel
7.85
21000
11.7
D2
7.85
21000
11.7
Powdered high-speed
steel（HAP40）
8.07
23300
10.1
Carbide V30
a
r
h
V＝ 3（A＋a＋ Aa）
│
Zone of sphere
r
Sphere
4
V＝ 3 πab 2
Volume V
π
V＝ 3 r2h
＝1.0472r2h
Truncated pyramid
In the case of a
rotating ellipsoidal
body（ b＝c）:
a
π
V＝ 4 h（D2－d2）
D
a is the radius.
4
V＝ 3 πabc
t
Cone
h
d
πh
＝ 6（3a 2＋h 2）
Ellipsoidal body
14.1
Cast iron
7.3
56000
7500 ∼ 10500
6.0
9.2 ∼ 11.8
304
8.0
19700
17.3
Oxygen-free
copper C1020
8.9
11700
17.6
6/4 brass C2801
8.4
10300
20.8
Aluminum A1100
2.7
6900
23.6
Duralumin A7075
2.8
7200
23.6
Titanium
4.5
10600
8.4
1kgf/mm2 ＝9.80665×106 Pa
Do not use in such a way that load is applied
to the threads.
■Finding the weight
■Finding dimensional changes resulting from thermal expansion ■Finding dimensional changes resulting from Young´s modulus E
Weight W［g］
×Density
＝Volume［cm3］
Example: Material: D2
Material: Soft steel
［Example］
Weight when D＝φ16 and L＝ 50 mm is found as follows.
L
PD-11.indd
＝19.739Rr2
π2
＝ 4 Dd2
ℓ
Hollow cylinder
（tube）
a
Root diameter d1
The information provided here is only an example of calculating the strength. For actual selections, it is necessary to consider the hole pitch accuracy, hole
perpendicularity, surface roughness, true roundness, plate material, parallelism, use of hardening, accuracy of the press machine, product production
volume, tool wear, and various other conditions. Therefore the strength calculation value should be used only as a guide.（It is not a guaranteed value.）
1271
d
π 2
V＝ 4 d（ℓ＋ℓ
´ 3）
D
πh 2
V＝ 3 （3r－h）
Fatigue strengths＊ have been excerpted from “Estimated values of fatigue limits
for metal threads of small screws, bolts, and nuts”（Yamamoto）and modified.
52100 yield stress capabilityσb＝120［kgf/mm2］
Maximum allowable shear strengthτ＝σb×0.8/（Safety factorα）
＝120×0.8/5
＝19.2［kgf/mm2］
F or an MS dowel pin, select a size of D8 or larger.
In addition, selecting a single size for all dowel pins makes it possible to reduce
items such as tools and inventory.
V＝2π2Rr2
R
h
Spherical crown
Barrel shape
≒7.3
Circular ring
d
h1
h
a
r
Strength class
Effective
Nominal cross-section
12.9
10.9
thread area
Maximum
Maximum
＊
Fatigue strength＊ allowable
load Fatigue strength allowable load
As
size
mm2 kgf /mm2
kgf
kgf /mm2
kgf
M 4
8.78 13.1
114
9.1
79
M 5
14.2
11.3
160
7.8
111
M 6
20.1
10.6
213
7.4
149
M 8
36.6
8.9
326
8.7
318
M10
58
7.4
429
7.3
423
M12
84.3
6.7
565
6.5
548
M14 115
6.1
702
6
690
M16 157
5.8
911
5.7
895
M20 245
5.2
1274
5.1
1250
M24 353
4.7
1659
4.7
1659
Find a suitable size for a single dowel pin which is subjected to repeated
（pulsating）
shear loads of 800 kgf.（Dowel pin material: 52100 hardness 58 HRC or higher）
＝πD2τ/4
a
Crossing cylinders
h
h
V＝ 3 A＝ 6 arn
A＝Bottom surface area
r＝Radius of inscribed circle
a＝Length of 1 side of regular polygon
n＝Number of regular polygon sides
＝2.0944r2h
（For threads: fatigue strength ＝ count of 2 million）
■Bolt fatigue strength
■Dowel pin strength
P＝A ×τ
h2
Pyramid
2
2
V＝ 3 πr2h
D
W＝π
－ D2× L×Density
4
π
＝－ ×1.62×5×7.85
4
≒79［g］
Example: Find strainλwhen load P＝1000 kgf is applied
to aφ10×L60 pin.（Material: D2）
Example: The amount of dimensional change δ
which occurs when a pin of D＝φ2, L＝100 mm
PL
is heated to 100℃ is the following.
E＝
Aλ
δ＝Coefficient of thermal expansion×Total length×Temperature change
1000×60
PL
λ＝
＝ 78.5×21000
＝11.7×10−6×100 mm×100℃
AE
＝0.117［mm］
≒0.036mm
100
［20℃］
π
Cross-section area A＝ 4 D2 ＝78.5
0.117
φ10
λ
Find the maximum allowable load P when a MSW30 screw plug is subjected to impact load.
（MSW30 material: 1045, tensile strengthσb at 34～43 HRC 65 kgf/mm2）
Assuming fracture due to shear
Shear cross-section area A＝Root diameter d1×π×L
occurs at the MSW root diameter
（Root diameter d1≒M－P）
location, the maximum allowable A＝（M－P）πL＝（30－1.5）π×12
load P＝τt×A.
＝1074［mm2］
＝3.9×107.4
Yield stress≒0.9×Tensile strengthσb＝0.9×65＝58.2
＝4190
［kgf］
Shear stress≒0.8×Yield stress
＝46.6
When the tap is a soft materials, Maximum allowable shear stressτt＝Shear stress /（Safety factor 12）
find the maximum allowable
＝46.6/12＝3.9［kgf/mm2］
shear from the inside thread
root diameter.
d
）
a2＋b2
π2
V＝ 4 d2
2
Pkgf
L
■Screw plug strength
（
h 1＋h 2
π
＝ 4 d2
2
For brittle materials＝Fracture stress
Volume V
Conical section of
sphere
h
From the table at right, for a strength class of 10.9 and a maximum allowable
load of 200 kgf, the suitable bolt is a 318［kgf］
M8. Therefore we select a 10 mm
MSB10 with a M8 thread section. When the bolt is subjected to shear load, also
use a dowel pin.
3D shape
φ2
2）
For stripper bolts and others which are subjected to tensile impact loads, the
selection is made based on the fatigue strength.（The bolt is subjected to 200
kgf loads in the same way. Stripper bolt material: 4137 33～38 HRC, strength
class 10.9.）
π
V＝ 4 d 2h
Yield stress for strength class 12.9σb＝112［kgf/mm ］
Maximum allowable stressσt＝σb/
（safety factor）
（From table above, safety factor＝5）
＝112/5
＝22.4［kgf/mm2］
Volume V
Ellipsoidal ring
b
Example: Find a suitable size for a single hexagon socket head cap screw that will
be subjected to repeated（pulsating）tensile loads of P＝200 kgf.（Hexagon socket
head cap screw material: 4137, 38～43 HRC, strength class 12.9）
From formula（1）:
As＝Pt/σt
＝200/22.4
＝8.9［mm2］
Finding the effective cross-section area
larger than this value from the table at
right shows that a 14.2［mm2］
M5 cap
screw should be selected.
With additional consideration for the fatigue strength, and based on the
strength class of 12.9 in the table, we select an M6 screw with maximum
allowable load of 213 kgf.
Safety factorα
3D shape
ℓ´
Volume V
Truncated cylinder
h
Pt＝σt×As……（1）
＝πd2σt/4… （2）
3D shape
h
1）
When bolt is subjected to tensile load
Pt : Tensile load in axial direction［kgf］ ■Unwin safety factorαbased on tensile strength
Repeated load
σb: Bolt yield stress［kgf/mm2］
Static
Impact
R
load Pulsating Alternating load
σt: Bolt maximum allowable stress
2
［kgf/mm ］
Steel
3
5
8
12
（σt＝σb/（safety factorα））
Cast iron
4
6
10
15
2
As : Bolt effective cross-section area［mm ］
Copper, soft metals
5
5
9
15
As＝πd2/4
d : Bolt effective diameter（root diameter）
［mm］ Shear stress＝ Standard strength Standard strength: F or ductile materials＝Yield stress
［TECHNICAL DATA］CALCULATION OF CUBIC VOLUME AND MATERIAL PHYSICAL PROPERTIES
P
■Bolt strength
OF BOLTS, SCREW PLUGS, AND DOWEL PINS
h
［TECHNICAL DATA］STRENGTH
［120℃］
1272
2012/10/25
17:14:11